A metal wire of resistance  3\:\Omega is elongated to make a uniform wire of double its previous length . This new wire is now bent and the ends joined to make a circle . If two points on this circle make an angle at the centre , the equivalent resistance between these two points will be :

  • Option 1)

    \frac{12}{5}\Omega

  • Option 2)

    \frac{5}{12}\Omega

     

  • Option 3)

    \frac{5}{3}\Omega

  • Option 4)

    \frac{7}{2}\Omega

 

Answers (1)

R_{initial }=3\: \Omega

After elongation

Vol of wire initial = vol of wire  final

\Rightarrow\pi\: r_{i}^{2}l_{i}=\pi r_{j}^{2}\:2\:l_{i}

\Rightarrow \:\:r_{j}^{2}=\frac{r_{i}^{2}}{2}

R_{i}=\rho \:\frac{l}{A}=\rho \:\frac{l_{i}}{\pi\:r_{i}^{2}}

R_{final}(after \:\:elongation)=\rho \:\frac{2l_{i}}{\pi\frac{r_{i}^{2}}{2}}=4Ri=12 \Omega

R_{1}= \frac{\left ( \frac{5\pi}{3} \right )}{2\pi}\times 12= 10 \Omega

R_{2}= \frac{\left ( \frac{\pi}{3} \right )}{2\pi}\times 12= 2 \Omega

R_{eq}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}=\frac{5}{3} \Omega


Option 1)

\frac{12}{5}\Omega

Option 2)

\frac{5}{12}\Omega

 

Option 3)

\frac{5}{3}\Omega

Option 4)

\frac{7}{2}\Omega

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